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Prove that the limit $$\lim_{x \to \infty} (x/(x+1))(\sin(x^2)) = \lim_{x \to \infty} \frac{x \sin x^2 }{x+1}$$ does not exist as x approaches positive infinity

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Hint: $x/(x+1)$ tends to $1$ as $x \to \infty$. What happens with the other factor? – mrf Oct 31 '12 at 21:03
Hint: consider the facts that (a) $\sin x$ will take values +1 and -1 infinitely often, and (b) the fraction at the front approaches a fixed value. – Old John Oct 31 '12 at 21:04
up vote 3 down vote accepted

It is sufficient to choose two sequences $\left\lbrace {x_n^{(1)}}\right\rbrace$ and $\left\lbrace {x_n^{(2)}}\right\rbrace$ such that $$\lim\limits_{n \rightarrow \infty} x_n^{(1)}=+\infty, \quad \lim\limits_{n \rightarrow \infty} x_n^{(2)}=+\infty $$ and $$\lim\limits_{n \rightarrow \infty} \sin{\left(x_n^{(1)}\right)^2} \ne \lim\limits_{n \rightarrow \infty} \sin{\left(x_n^{(2)}\right)^2}$$ what means that $\lim\limits_{x \rightarrow +\infty} \sin{x^{2}}$ does not exist.

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Note that $$\lim_{x \to \infty} \dfrac{x}{x+1} \sin(x^2) =\left(\lim_{x \to \infty} \dfrac{x}{x+1} \right) \left(\lim_{x \to \infty} \sin(x^2) \right)$$

What can you say about $\lim_{x \to \infty} \dfrac{x}{x+1}$ and $\lim_{x \to \infty} \sin(x^2)$? (The first term is nice and converges whereas the second one is the problematic one).

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I'm not sure equality can be used in case of divergent limits... – gt6989b Oct 31 '12 at 21:39

Hint Make the change of variables $x=\frac{1}{y}$ (now you are considering the limit of the function as $y \to 0 $) and pick up the sequence $ {\frac{\sqrt{2}}{\sqrt{(2n+1)\pi}}} $ which goes to zero as $n\to \infty$ and see the values the function achieves as $n \to \infty$.

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